Question

A simple atwood's machine uses two masses, m1 and m2. starting from rest, the speed of the two masses is 2 m/s at the end of 5 s. at that instant, the kinetic energy of the system is 87 j and each mass has moved a distance of 5 m. determine the values of m1 and m2.

Answer 1

In Atwood machine system we will have

`a = (m_1 - m_2)/(m_1+m_2)*g`

`0.4 = (m_1 - m_2)/(m_1 + m_2)*10`

`0.4m_1 + 0.4m_2 = 10 m_1 - 10m_2`

`10.4m_2 = 9.6m_1`

`1.08m_2 = m_1`

Also we know the kinetic energy

`(1)/(2) m_1v_1^2 +(1)/(2)m_2v_2^2 = 87`

`m_1 + m_2 = 43.5 J`

now solving the two equations we will have

`m_2 = 21 kg`

`m_1 = 22.6 kg`